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CHXII09:COORDINATION COMPOUNDS

322298 Assertion :
${[Ni (CN)_{4}]}^{2 -}$ has square planar and
${[{NiCl}_{4}]}^{2 -}$ has tetrahedral shape.
Reason :
${[Ni (CN)_{4}]}^{2 -}$ is diamagnetic while
${[{NiCl}_{4}]}^{2 -}$ is paramagnetic.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

(i) Formation of ${[{NiCl}_{4}]}^{2 -}$
supporting img
It is tetrahedral due to ${sp}^{3}$ hybridization and paramagnetic due to two unpaired electrons.
(i) Formation of ${[{NiCl}_{4}]}^{2 -}$

It is square planar due to ${dsp}^{2}$ hybridization. and is diamagnetic as all electrons are paired. So the option (2) is correct.

CHXII09:COORDINATION COMPOUNDS

322299 Assertion :
$K_{4} [Fe (CN)_{6}]$ is diamagnetic and $[Fe {(H_{2} O)}_{6}] {Cl}_{3}$ is paramagnetic.
Reason :
Hybridisation of central metal in $K_{4} [Fe (CN)_{6}]$ is ${sp}^{3} d^{2}$ , while in $[Fe {(H_{2} O)}_{6}] {Cl}_{3}$ is $d^{2} {sp}^{3}$ .

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and reason is incorrect.

Explanation:

${In}_{4} [Fe (CN)_{6}]$ , $Fe$ is present as ${Fe}^{2 +}$ $Fe = [Ar] 3 d^{6} 4 s^{2}$
${Fe}^{2 +} = [Ar] 3 d^{6}, 4 s^{0}$
supporting img
Since, all electorns are paired, it is diamagnetic in nature.
$In [Fe {(H_{2} O)}_{6}] {Cl}_{3}, Fe$ is present as ${Fe}^{3 +}$
${Fe}^{3 +} = [Ar] 3 d^{5}, 4 s^{0}$
${[Fe {(H_{2} O)}_{6}]}^{3 +} =$

Five unpaired electrons ${sp}^{3} d^{2}$
Since, it contains five unpaired electrons, it is paramagnetic in nature.

AIIMS - 2009

CHXII09:COORDINATION COMPOUNDS

322300 Assertion :
$[Fe {(H_{2} O)}_{5} {NO}_{3} {SO}_{4}$ is
paramagnetic.
Reason :
The $Fe$ in $[Fe {(H_{2} O)}_{5} {NO}_{3} {SO}_{4}$ has three unpaired electrons.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

${Fe}^{+} : [Ar] 3 d^{6} 4 s^{1}$
When both the weak field ligand $H_{2} O$ and strong field ligand $NO$ attack, the configuration changes as follows:
supporting img
${Fe}^{+}$ has 3 unpaired electrons. So the option (1) is correct.

CHXII09:COORDINATION COMPOUNDS

322301 Nickel $(Z = 28)$ combines with a uninegative monodentate ligand $X^{-}$ to form a paramagnetic complex ${[{NiX}_{4}]}^{2 -}$ . The number of unpaired electrons in the nickel and geometry of this complex ion are, respectively

1 One and tetrahedral

2 Two and tetrahedral

3 One and square planar

4 Two and square planar

Explanation:

${[{NiX}_{4}]}^{2 -}$ is paramagnetic complex.
So, ${Ni}^{+ 2}$ will have $d^{8}$ configuration with 2 unpaired electrons. It undergoes ${sp}^{3}$ hybridization and has tetrahedral geometry.

CHXII09:COORDINATION COMPOUNDS

322302 Which of the following is not expected to show paramagnetism?

1

{[Ni (H_{2} O)]}^{2 +}

2

Ni (CO)_{4}

3

{[Ni {({NH}_{3})}_{4}]}^{2 +}

4

{[Co {({NH}_{3})}_{6}]}^{2 +}

Explanation:

$Ni (CO)_{4}$ due to absence of unpaired electrons does not show paramagnetism.
$N i \to {[A r]}_{18} 3 d^{8} 4 s^{2}$

CHXII09:COORDINATION COMPOUNDS

322298 Assertion :
${[Ni (CN)_{4}]}^{2 -}$ has square planar and
${[{NiCl}_{4}]}^{2 -}$ has tetrahedral shape.
Reason :
${[Ni (CN)_{4}]}^{2 -}$ is diamagnetic while
${[{NiCl}_{4}]}^{2 -}$ is paramagnetic.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

(i) Formation of ${[{NiCl}_{4}]}^{2 -}$
supporting img
It is tetrahedral due to ${sp}^{3}$ hybridization and paramagnetic due to two unpaired electrons.
(i) Formation of ${[{NiCl}_{4}]}^{2 -}$

It is square planar due to ${dsp}^{2}$ hybridization. and is diamagnetic as all electrons are paired. So the option (2) is correct.

CHXII09:COORDINATION COMPOUNDS

322299 Assertion :
$K_{4} [Fe (CN)_{6}]$ is diamagnetic and $[Fe {(H_{2} O)}_{6}] {Cl}_{3}$ is paramagnetic.
Reason :
Hybridisation of central metal in $K_{4} [Fe (CN)_{6}]$ is ${sp}^{3} d^{2}$ , while in $[Fe {(H_{2} O)}_{6}] {Cl}_{3}$ is $d^{2} {sp}^{3}$ .

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and reason is incorrect.

Explanation:

${In}_{4} [Fe (CN)_{6}]$ , $Fe$ is present as ${Fe}^{2 +}$ $Fe = [Ar] 3 d^{6} 4 s^{2}$
${Fe}^{2 +} = [Ar] 3 d^{6}, 4 s^{0}$
supporting img
Since, all electorns are paired, it is diamagnetic in nature.
$In [Fe {(H_{2} O)}_{6}] {Cl}_{3}, Fe$ is present as ${Fe}^{3 +}$
${Fe}^{3 +} = [Ar] 3 d^{5}, 4 s^{0}$
${[Fe {(H_{2} O)}_{6}]}^{3 +} =$

Five unpaired electrons ${sp}^{3} d^{2}$
Since, it contains five unpaired electrons, it is paramagnetic in nature.

AIIMS - 2009

CHXII09:COORDINATION COMPOUNDS

322300 Assertion :
$[Fe {(H_{2} O)}_{5} {NO}_{3} {SO}_{4}$ is
paramagnetic.
Reason :
The $Fe$ in $[Fe {(H_{2} O)}_{5} {NO}_{3} {SO}_{4}$ has three unpaired electrons.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

${Fe}^{+} : [Ar] 3 d^{6} 4 s^{1}$
When both the weak field ligand $H_{2} O$ and strong field ligand $NO$ attack, the configuration changes as follows:
supporting img
${Fe}^{+}$ has 3 unpaired electrons. So the option (1) is correct.

CHXII09:COORDINATION COMPOUNDS

322301 Nickel $(Z = 28)$ combines with a uninegative monodentate ligand $X^{-}$ to form a paramagnetic complex ${[{NiX}_{4}]}^{2 -}$ . The number of unpaired electrons in the nickel and geometry of this complex ion are, respectively

1 One and tetrahedral

2 Two and tetrahedral

3 One and square planar

4 Two and square planar

Explanation:

${[{NiX}_{4}]}^{2 -}$ is paramagnetic complex.
So, ${Ni}^{+ 2}$ will have $d^{8}$ configuration with 2 unpaired electrons. It undergoes ${sp}^{3}$ hybridization and has tetrahedral geometry.

CHXII09:COORDINATION COMPOUNDS

322302 Which of the following is not expected to show paramagnetism?

1

{[Ni (H_{2} O)]}^{2 +}

2

Ni (CO)_{4}

3

{[Ni {({NH}_{3})}_{4}]}^{2 +}

4

{[Co {({NH}_{3})}_{6}]}^{2 +}

Explanation:

$Ni (CO)_{4}$ due to absence of unpaired electrons does not show paramagnetism.
$N i \to {[A r]}_{18} 3 d^{8} 4 s^{2}$

CHXII09:COORDINATION COMPOUNDS

322298 Assertion :
${[Ni (CN)_{4}]}^{2 -}$ has square planar and
${[{NiCl}_{4}]}^{2 -}$ has tetrahedral shape.
Reason :
${[Ni (CN)_{4}]}^{2 -}$ is diamagnetic while
${[{NiCl}_{4}]}^{2 -}$ is paramagnetic.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

(i) Formation of ${[{NiCl}_{4}]}^{2 -}$
supporting img
It is tetrahedral due to ${sp}^{3}$ hybridization and paramagnetic due to two unpaired electrons.
(i) Formation of ${[{NiCl}_{4}]}^{2 -}$

It is square planar due to ${dsp}^{2}$ hybridization. and is diamagnetic as all electrons are paired. So the option (2) is correct.

CHXII09:COORDINATION COMPOUNDS

322299 Assertion :
$K_{4} [Fe (CN)_{6}]$ is diamagnetic and $[Fe {(H_{2} O)}_{6}] {Cl}_{3}$ is paramagnetic.
Reason :
Hybridisation of central metal in $K_{4} [Fe (CN)_{6}]$ is ${sp}^{3} d^{2}$ , while in $[Fe {(H_{2} O)}_{6}] {Cl}_{3}$ is $d^{2} {sp}^{3}$ .

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and reason is incorrect.

Explanation:

${In}_{4} [Fe (CN)_{6}]$ , $Fe$ is present as ${Fe}^{2 +}$ $Fe = [Ar] 3 d^{6} 4 s^{2}$
${Fe}^{2 +} = [Ar] 3 d^{6}, 4 s^{0}$
supporting img
Since, all electorns are paired, it is diamagnetic in nature.
$In [Fe {(H_{2} O)}_{6}] {Cl}_{3}, Fe$ is present as ${Fe}^{3 +}$
${Fe}^{3 +} = [Ar] 3 d^{5}, 4 s^{0}$
${[Fe {(H_{2} O)}_{6}]}^{3 +} =$

Five unpaired electrons ${sp}^{3} d^{2}$
Since, it contains five unpaired electrons, it is paramagnetic in nature.

AIIMS - 2009

CHXII09:COORDINATION COMPOUNDS

322300 Assertion :
$[Fe {(H_{2} O)}_{5} {NO}_{3} {SO}_{4}$ is
paramagnetic.
Reason :
The $Fe$ in $[Fe {(H_{2} O)}_{5} {NO}_{3} {SO}_{4}$ has three unpaired electrons.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

${Fe}^{+} : [Ar] 3 d^{6} 4 s^{1}$
When both the weak field ligand $H_{2} O$ and strong field ligand $NO$ attack, the configuration changes as follows:
supporting img
${Fe}^{+}$ has 3 unpaired electrons. So the option (1) is correct.

CHXII09:COORDINATION COMPOUNDS

322301 Nickel $(Z = 28)$ combines with a uninegative monodentate ligand $X^{-}$ to form a paramagnetic complex ${[{NiX}_{4}]}^{2 -}$ . The number of unpaired electrons in the nickel and geometry of this complex ion are, respectively

1 One and tetrahedral

2 Two and tetrahedral

3 One and square planar

4 Two and square planar

Explanation:

${[{NiX}_{4}]}^{2 -}$ is paramagnetic complex.
So, ${Ni}^{+ 2}$ will have $d^{8}$ configuration with 2 unpaired electrons. It undergoes ${sp}^{3}$ hybridization and has tetrahedral geometry.

CHXII09:COORDINATION COMPOUNDS

322302 Which of the following is not expected to show paramagnetism?

1

{[Ni (H_{2} O)]}^{2 +}

2

Ni (CO)_{4}

3

{[Ni {({NH}_{3})}_{4}]}^{2 +}

4

{[Co {({NH}_{3})}_{6}]}^{2 +}

Explanation:

$Ni (CO)_{4}$ due to absence of unpaired electrons does not show paramagnetism.
$N i \to {[A r]}_{18} 3 d^{8} 4 s^{2}$

CHXII09:COORDINATION COMPOUNDS

322298 Assertion :
${[Ni (CN)_{4}]}^{2 -}$ has square planar and
${[{NiCl}_{4}]}^{2 -}$ has tetrahedral shape.
Reason :
${[Ni (CN)_{4}]}^{2 -}$ is diamagnetic while
${[{NiCl}_{4}]}^{2 -}$ is paramagnetic.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

(i) Formation of ${[{NiCl}_{4}]}^{2 -}$
supporting img
It is tetrahedral due to ${sp}^{3}$ hybridization and paramagnetic due to two unpaired electrons.
(i) Formation of ${[{NiCl}_{4}]}^{2 -}$

It is square planar due to ${dsp}^{2}$ hybridization. and is diamagnetic as all electrons are paired. So the option (2) is correct.

CHXII09:COORDINATION COMPOUNDS

322299 Assertion :
$K_{4} [Fe (CN)_{6}]$ is diamagnetic and $[Fe {(H_{2} O)}_{6}] {Cl}_{3}$ is paramagnetic.
Reason :
Hybridisation of central metal in $K_{4} [Fe (CN)_{6}]$ is ${sp}^{3} d^{2}$ , while in $[Fe {(H_{2} O)}_{6}] {Cl}_{3}$ is $d^{2} {sp}^{3}$ .

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and reason is incorrect.

Explanation:

${In}_{4} [Fe (CN)_{6}]$ , $Fe$ is present as ${Fe}^{2 +}$ $Fe = [Ar] 3 d^{6} 4 s^{2}$
${Fe}^{2 +} = [Ar] 3 d^{6}, 4 s^{0}$
supporting img
Since, all electorns are paired, it is diamagnetic in nature.
$In [Fe {(H_{2} O)}_{6}] {Cl}_{3}, Fe$ is present as ${Fe}^{3 +}$
${Fe}^{3 +} = [Ar] 3 d^{5}, 4 s^{0}$
${[Fe {(H_{2} O)}_{6}]}^{3 +} =$

Five unpaired electrons ${sp}^{3} d^{2}$
Since, it contains five unpaired electrons, it is paramagnetic in nature.

AIIMS - 2009

CHXII09:COORDINATION COMPOUNDS

322300 Assertion :
$[Fe {(H_{2} O)}_{5} {NO}_{3} {SO}_{4}$ is
paramagnetic.
Reason :
The $Fe$ in $[Fe {(H_{2} O)}_{5} {NO}_{3} {SO}_{4}$ has three unpaired electrons.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

${Fe}^{+} : [Ar] 3 d^{6} 4 s^{1}$
When both the weak field ligand $H_{2} O$ and strong field ligand $NO$ attack, the configuration changes as follows:
supporting img
${Fe}^{+}$ has 3 unpaired electrons. So the option (1) is correct.

CHXII09:COORDINATION COMPOUNDS

322301 Nickel $(Z = 28)$ combines with a uninegative monodentate ligand $X^{-}$ to form a paramagnetic complex ${[{NiX}_{4}]}^{2 -}$ . The number of unpaired electrons in the nickel and geometry of this complex ion are, respectively

1 One and tetrahedral

2 Two and tetrahedral

3 One and square planar

4 Two and square planar

Explanation:

${[{NiX}_{4}]}^{2 -}$ is paramagnetic complex.
So, ${Ni}^{+ 2}$ will have $d^{8}$ configuration with 2 unpaired electrons. It undergoes ${sp}^{3}$ hybridization and has tetrahedral geometry.

CHXII09:COORDINATION COMPOUNDS

322302 Which of the following is not expected to show paramagnetism?

1

{[Ni (H_{2} O)]}^{2 +}

2

Ni (CO)_{4}

3

{[Ni {({NH}_{3})}_{4}]}^{2 +}

4

{[Co {({NH}_{3})}_{6}]}^{2 +}

Explanation:

$Ni (CO)_{4}$ due to absence of unpaired electrons does not show paramagnetism.
$N i \to {[A r]}_{18} 3 d^{8} 4 s^{2}$

CHXII09:COORDINATION COMPOUNDS

322298 Assertion :
${[Ni (CN)_{4}]}^{2 -}$ has square planar and
${[{NiCl}_{4}]}^{2 -}$ has tetrahedral shape.
Reason :
${[Ni (CN)_{4}]}^{2 -}$ is diamagnetic while
${[{NiCl}_{4}]}^{2 -}$ is paramagnetic.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

(i) Formation of ${[{NiCl}_{4}]}^{2 -}$
supporting img
It is tetrahedral due to ${sp}^{3}$ hybridization and paramagnetic due to two unpaired electrons.
(i) Formation of ${[{NiCl}_{4}]}^{2 -}$

It is square planar due to ${dsp}^{2}$ hybridization. and is diamagnetic as all electrons are paired. So the option (2) is correct.

CHXII09:COORDINATION COMPOUNDS

322299 Assertion :
$K_{4} [Fe (CN)_{6}]$ is diamagnetic and $[Fe {(H_{2} O)}_{6}] {Cl}_{3}$ is paramagnetic.
Reason :
Hybridisation of central metal in $K_{4} [Fe (CN)_{6}]$ is ${sp}^{3} d^{2}$ , while in $[Fe {(H_{2} O)}_{6}] {Cl}_{3}$ is $d^{2} {sp}^{3}$ .

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and reason is incorrect.

Explanation:

${In}_{4} [Fe (CN)_{6}]$ , $Fe$ is present as ${Fe}^{2 +}$ $Fe = [Ar] 3 d^{6} 4 s^{2}$
${Fe}^{2 +} = [Ar] 3 d^{6}, 4 s^{0}$
supporting img
Since, all electorns are paired, it is diamagnetic in nature.
$In [Fe {(H_{2} O)}_{6}] {Cl}_{3}, Fe$ is present as ${Fe}^{3 +}$
${Fe}^{3 +} = [Ar] 3 d^{5}, 4 s^{0}$
${[Fe {(H_{2} O)}_{6}]}^{3 +} =$

Five unpaired electrons ${sp}^{3} d^{2}$
Since, it contains five unpaired electrons, it is paramagnetic in nature.

AIIMS - 2009

CHXII09:COORDINATION COMPOUNDS

322300 Assertion :
$[Fe {(H_{2} O)}_{5} {NO}_{3} {SO}_{4}$ is
paramagnetic.
Reason :
The $Fe$ in $[Fe {(H_{2} O)}_{5} {NO}_{3} {SO}_{4}$ has three unpaired electrons.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

${Fe}^{+} : [Ar] 3 d^{6} 4 s^{1}$
When both the weak field ligand $H_{2} O$ and strong field ligand $NO$ attack, the configuration changes as follows:
supporting img
${Fe}^{+}$ has 3 unpaired electrons. So the option (1) is correct.

CHXII09:COORDINATION COMPOUNDS

322301 Nickel $(Z = 28)$ combines with a uninegative monodentate ligand $X^{-}$ to form a paramagnetic complex ${[{NiX}_{4}]}^{2 -}$ . The number of unpaired electrons in the nickel and geometry of this complex ion are, respectively

1 One and tetrahedral

2 Two and tetrahedral

3 One and square planar

4 Two and square planar

Explanation:

${[{NiX}_{4}]}^{2 -}$ is paramagnetic complex.
So, ${Ni}^{+ 2}$ will have $d^{8}$ configuration with 2 unpaired electrons. It undergoes ${sp}^{3}$ hybridization and has tetrahedral geometry.

CHXII09:COORDINATION COMPOUNDS

322302 Which of the following is not expected to show paramagnetism?

1

{[Ni (H_{2} O)]}^{2 +}

2

Ni (CO)_{4}

3

{[Ni {({NH}_{3})}_{4}]}^{2 +}

4

{[Co {({NH}_{3})}_{6}]}^{2 +}

Explanation:

$Ni (CO)_{4}$ due to absence of unpaired electrons does not show paramagnetism.
$N i \to {[A r]}_{18} 3 d^{8} 4 s^{2}$

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